Simplify the following expression: $y = \dfrac{-3x^2- 17x- 10}{-3x - 2}$
First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-3)}{(-10)} &=& 30 \\ {a} + {b} &=& &=& {-17} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $30$ and add them together. The factors that add up to ${-17}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-2}$ and ${b}$ is ${-15}$ $ \begin{eqnarray} {ab} &=& ({-2})({-15}) &=& 30 \\ {a} + {b} &=& {-2} + {-15} &=& -17 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({-3}x^2 {-2}x) + ({-15}x {-10}) $ Factor out the common factors: $ x(-3x - 2) + 5(-3x - 2)$ Now factor out $(-3x - 2)$ $ (-3x - 2)(x + 5)$ The original expression can therefore be written: $ \dfrac{(-3x - 2)(x + 5)}{-3x - 2}$ We are dividing by $-3x - 2$ , so $-3x - 2 \neq 0$ Therefore, $x \neq -\frac{2}{3}$ This leaves us with $x + 5; x \neq -\frac{2}{3}$.